Question

$\underset{n\to \infty }{lim}(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}......\frac{1}{5n})$

• A. $\pi /4$
• B. $\pi /2$
• C. ${\tan }^{-1}\left(2\right)$
• D. ${\tan }^{-1}$

Answer 1

Answer: (C)

${\tan }^{-1}\left(2\right)$

Let

$S={lim}_{n\to \infty }[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+....+\frac{n}{n^2+{\left(2n\right)}^2}]$

$S={lim}_{n\to \infty }\frac{1}{n}[\frac{n^2/1^2}{\frac{n^2}{1^2}+1}+\frac{n^2/2^2}{\frac{n^2}{2^2}+1}+.....+\frac{n^2/{\left(2n\right)}^2}{\frac{n^2}{{\left(2n\right)}^2}+1}]$

$S={lim}_{n\to \infty }\left[\frac{1}{n}{\sum }_{r=1}^{2n}\frac{n^2/r^2}{\frac{n^2}{r^2}+1}\right]$

Let $\frac{r}{n}=x$

So, at $r=1,x=0$

and at $r=2n,x=2$

So, $S={\int }_0^2\frac{1/x^2}{\frac{1}{x^2}+1}dx={\int }_0^2\frac{1}{1+x^2}dx$

$S={\tan }^{-1}\left(x\right){/}_0^2={\tan }^{-1}\left(2\right)$